Internal problem ID [10477]
Internal file name [OUTPUT/9425_Monday_June_06_2022_02_31_57_PM_8846351/index.tex]
Book: Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev.
Second edition
Section: Chapter 1, section 1.2. Riccati Equation. subsection 1.2.5-1. Equations Containing
Logarithmic Functions
Problem number: 3.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "riccati"
Maple gives the following as the ode type
[_Riccati]
\[ \boxed {y^{\prime } x -a y^{2}=b \ln \left (x \right )^{k}+c \ln \left (x \right )^{2 k +2}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {a \,y^{2}+b \ln \left (x \right )^{k}+c \ln \left (x \right )^{2 k +2}}{x} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = \frac {a \,y^{2}}{x}+\frac {b \ln \left (x \right )^{k}}{x}+\frac {c \ln \left (x \right )^{2 k} \ln \left (x \right )^{2}}{x} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\frac {b \ln \left (x \right )^{k}+c \ln \left (x \right )^{2 k +2}}{x}\), \(f_1(x)=0\) and \(f_2(x)=\frac {a}{x}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\frac {a u}{x}} \tag {1} \end {align*}
Using the above substitution in the given ODE results (after some simplification)in a second order ODE to solve for \(u(x)\) which is \begin {align*} f_2 u''(x) -\left ( f_2' + f_1 f_2 \right ) u'(x) + f_2^2 f_0 u(x) &= 0 \tag {2} \end {align*}
But \begin {align*} f_2' &=-\frac {a}{x^{2}}\\ f_1 f_2 &=0\\ f_2^2 f_0 &=\frac {a^{2} \left (b \ln \left (x \right )^{k}+c \ln \left (x \right )^{2 k +2}\right )}{x^{3}} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} \frac {a u^{\prime \prime }\left (x \right )}{x}+\frac {a u^{\prime }\left (x \right )}{x^{2}}+\frac {a^{2} \left (b \ln \left (x \right )^{k}+c \ln \left (x \right )^{2 k +2}\right ) u \left (x \right )}{x^{3}} &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = {\mathrm e}^{-\frac {i \sqrt {c}\, \sqrt {a}\, \ln \left (x \right )^{2+k}}{2+k}} \left (\operatorname {hypergeom}\left (\left [\frac {\left (k +1\right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (4+2 k \right )}\right ], \left [\frac {k +1}{2+k}\right ], \frac {2 i \sqrt {c}\, \sqrt {a}\, \ln \left (x \right )^{2+k}}{2+k}\right ) c_{1} +\operatorname {hypergeom}\left (\left [\frac {\left (k +3\right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (4+2 k \right )}\right ], \left [\frac {k +3}{2+k}\right ], \frac {2 i \sqrt {c}\, \sqrt {a}\, \ln \left (x \right )^{2+k}}{2+k}\right ) c_{2} \ln \left (x \right )\right ) \] The above shows that \[ u^{\prime }\left (x \right ) = -\frac {\left (-\left (k +3\right ) c_{1} \left (i \left (k +1\right ) \sqrt {c}\, \sqrt {a}-a b \right ) \ln \left (x \right )^{k +1} \operatorname {hypergeom}\left (\left [\frac {\left (5+3 k \right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (4+2 k \right )}\right ], \left [\frac {3+2 k}{2+k}\right ], \frac {2 i \sqrt {c}\, \sqrt {a}\, \ln \left (x \right )^{2+k}}{2+k}\right )+\left (k +1\right ) \left (-\ln \left (x \right )^{2+k} \left (i \sqrt {c}\, \left (k +3\right ) \sqrt {a}-a b \right ) c_{2} \operatorname {hypergeom}\left (\left [\frac {\left (7+3 k \right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (4+2 k \right )}\right ], \left [\frac {5+2 k}{2+k}\right ], \frac {2 i \sqrt {c}\, \sqrt {a}\, \ln \left (x \right )^{2+k}}{2+k}\right )+\left (k +3\right ) \left (\left (i \sqrt {c}\, \sqrt {a}\, \ln \left (x \right )^{2+k}-1\right ) c_{2} \operatorname {hypergeom}\left (\left [\frac {\left (k +3\right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (4+2 k \right )}\right ], \left [\frac {k +3}{2+k}\right ], \frac {2 i \sqrt {c}\, \sqrt {a}\, \ln \left (x \right )^{2+k}}{2+k}\right )+i c_{1} \ln \left (x \right )^{k +1} \sqrt {a}\, \sqrt {c}\, \operatorname {hypergeom}\left (\left [\frac {\left (k +1\right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (4+2 k \right )}\right ], \left [\frac {k +1}{2+k}\right ], \frac {2 i \sqrt {c}\, \sqrt {a}\, \ln \left (x \right )^{2+k}}{2+k}\right )\right )\right )\right ) {\mathrm e}^{-\frac {i \sqrt {c}\, \sqrt {a}\, \ln \left (x \right )^{2+k}}{2+k}}}{\left (k +3\right ) x \left (k +1\right )} \] Using the above in (1) gives the solution \[ y = \frac {-\left (k +3\right ) c_{1} \left (i \left (k +1\right ) \sqrt {c}\, \sqrt {a}-a b \right ) \ln \left (x \right )^{k +1} \operatorname {hypergeom}\left (\left [\frac {\left (5+3 k \right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (4+2 k \right )}\right ], \left [\frac {3+2 k}{2+k}\right ], \frac {2 i \sqrt {c}\, \sqrt {a}\, \ln \left (x \right )^{2+k}}{2+k}\right )+\left (k +1\right ) \left (-\ln \left (x \right )^{2+k} \left (i \sqrt {c}\, \left (k +3\right ) \sqrt {a}-a b \right ) c_{2} \operatorname {hypergeom}\left (\left [\frac {\left (7+3 k \right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (4+2 k \right )}\right ], \left [\frac {5+2 k}{2+k}\right ], \frac {2 i \sqrt {c}\, \sqrt {a}\, \ln \left (x \right )^{2+k}}{2+k}\right )+\left (k +3\right ) \left (\left (i \sqrt {c}\, \sqrt {a}\, \ln \left (x \right )^{2+k}-1\right ) c_{2} \operatorname {hypergeom}\left (\left [\frac {\left (k +3\right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (4+2 k \right )}\right ], \left [\frac {k +3}{2+k}\right ], \frac {2 i \sqrt {c}\, \sqrt {a}\, \ln \left (x \right )^{2+k}}{2+k}\right )+i c_{1} \ln \left (x \right )^{k +1} \sqrt {a}\, \sqrt {c}\, \operatorname {hypergeom}\left (\left [\frac {\left (k +1\right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (4+2 k \right )}\right ], \left [\frac {k +1}{2+k}\right ], \frac {2 i \sqrt {c}\, \sqrt {a}\, \ln \left (x \right )^{2+k}}{2+k}\right )\right )\right )}{\left (k +3\right ) \left (k +1\right ) a \left (\operatorname {hypergeom}\left (\left [\frac {\left (k +1\right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (4+2 k \right )}\right ], \left [\frac {k +1}{2+k}\right ], \frac {2 i \sqrt {c}\, \sqrt {a}\, \ln \left (x \right )^{2+k}}{2+k}\right ) c_{1} +\operatorname {hypergeom}\left (\left [\frac {\left (k +3\right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (4+2 k \right )}\right ], \left [\frac {k +3}{2+k}\right ], \frac {2 i \sqrt {c}\, \sqrt {a}\, \ln \left (x \right )^{2+k}}{2+k}\right ) c_{2} \ln \left (x \right )\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {-\left (k +3\right ) c_{3} \left (i \left (k +1\right ) \sqrt {c}\, \sqrt {a}-a b \right ) \ln \left (x \right )^{k +1} \operatorname {hypergeom}\left (\left [\frac {\left (5+3 k \right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (4+2 k \right )}\right ], \left [\frac {3+2 k}{2+k}\right ], \frac {2 i \sqrt {c}\, \sqrt {a}\, \ln \left (x \right )^{2+k}}{2+k}\right )+\left (k +1\right ) \left (-\ln \left (x \right )^{2+k} \left (i \sqrt {c}\, \left (k +3\right ) \sqrt {a}-a b \right ) \operatorname {hypergeom}\left (\left [\frac {\left (7+3 k \right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (4+2 k \right )}\right ], \left [\frac {5+2 k}{2+k}\right ], \frac {2 i \sqrt {c}\, \sqrt {a}\, \ln \left (x \right )^{2+k}}{2+k}\right )+\left (k +3\right ) \left (\left (i \sqrt {c}\, \sqrt {a}\, \ln \left (x \right )^{2+k}-1\right ) \operatorname {hypergeom}\left (\left [\frac {\left (k +3\right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (4+2 k \right )}\right ], \left [\frac {k +3}{2+k}\right ], \frac {2 i \sqrt {c}\, \sqrt {a}\, \ln \left (x \right )^{2+k}}{2+k}\right )+i c_{3} \ln \left (x \right )^{k +1} \sqrt {a}\, \sqrt {c}\, \operatorname {hypergeom}\left (\left [\frac {\left (k +1\right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (4+2 k \right )}\right ], \left [\frac {k +1}{2+k}\right ], \frac {2 i \sqrt {c}\, \sqrt {a}\, \ln \left (x \right )^{2+k}}{2+k}\right )\right )\right )}{\left (k +3\right ) \left (k +1\right ) a \left (\operatorname {hypergeom}\left (\left [\frac {\left (k +1\right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (4+2 k \right )}\right ], \left [\frac {k +1}{2+k}\right ], \frac {2 i \sqrt {c}\, \sqrt {a}\, \ln \left (x \right )^{2+k}}{2+k}\right ) c_{3} +\operatorname {hypergeom}\left (\left [\frac {\left (k +3\right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (4+2 k \right )}\right ], \left [\frac {k +3}{2+k}\right ], \frac {2 i \sqrt {c}\, \sqrt {a}\, \ln \left (x \right )^{2+k}}{2+k}\right ) \ln \left (x \right )\right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {-\left (k +3\right ) c_{3} \left (i \left (k +1\right ) \sqrt {c}\, \sqrt {a}-a b \right ) \ln \left (x \right )^{k +1} \operatorname {hypergeom}\left (\left [\frac {\left (5+3 k \right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (4+2 k \right )}\right ], \left [\frac {3+2 k}{2+k}\right ], \frac {2 i \sqrt {c}\, \sqrt {a}\, \ln \left (x \right )^{2+k}}{2+k}\right )+\left (k +1\right ) \left (-\ln \left (x \right )^{2+k} \left (i \sqrt {c}\, \left (k +3\right ) \sqrt {a}-a b \right ) \operatorname {hypergeom}\left (\left [\frac {\left (7+3 k \right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (4+2 k \right )}\right ], \left [\frac {5+2 k}{2+k}\right ], \frac {2 i \sqrt {c}\, \sqrt {a}\, \ln \left (x \right )^{2+k}}{2+k}\right )+\left (k +3\right ) \left (\left (i \sqrt {c}\, \sqrt {a}\, \ln \left (x \right )^{2+k}-1\right ) \operatorname {hypergeom}\left (\left [\frac {\left (k +3\right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (4+2 k \right )}\right ], \left [\frac {k +3}{2+k}\right ], \frac {2 i \sqrt {c}\, \sqrt {a}\, \ln \left (x \right )^{2+k}}{2+k}\right )+i c_{3} \ln \left (x \right )^{k +1} \sqrt {a}\, \sqrt {c}\, \operatorname {hypergeom}\left (\left [\frac {\left (k +1\right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (4+2 k \right )}\right ], \left [\frac {k +1}{2+k}\right ], \frac {2 i \sqrt {c}\, \sqrt {a}\, \ln \left (x \right )^{2+k}}{2+k}\right )\right )\right )}{\left (k +3\right ) \left (k +1\right ) a \left (\operatorname {hypergeom}\left (\left [\frac {\left (k +1\right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (4+2 k \right )}\right ], \left [\frac {k +1}{2+k}\right ], \frac {2 i \sqrt {c}\, \sqrt {a}\, \ln \left (x \right )^{2+k}}{2+k}\right ) c_{3} +\operatorname {hypergeom}\left (\left [\frac {\left (k +3\right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (4+2 k \right )}\right ], \left [\frac {k +3}{2+k}\right ], \frac {2 i \sqrt {c}\, \sqrt {a}\, \ln \left (x \right )^{2+k}}{2+k}\right ) \ln \left (x \right )\right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {-\left (k +3\right ) c_{3} \left (i \left (k +1\right ) \sqrt {c}\, \sqrt {a}-a b \right ) \ln \left (x \right )^{k +1} \operatorname {hypergeom}\left (\left [\frac {\left (5+3 k \right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (4+2 k \right )}\right ], \left [\frac {3+2 k}{2+k}\right ], \frac {2 i \sqrt {c}\, \sqrt {a}\, \ln \left (x \right )^{2+k}}{2+k}\right )+\left (k +1\right ) \left (-\ln \left (x \right )^{2+k} \left (i \sqrt {c}\, \left (k +3\right ) \sqrt {a}-a b \right ) \operatorname {hypergeom}\left (\left [\frac {\left (7+3 k \right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (4+2 k \right )}\right ], \left [\frac {5+2 k}{2+k}\right ], \frac {2 i \sqrt {c}\, \sqrt {a}\, \ln \left (x \right )^{2+k}}{2+k}\right )+\left (k +3\right ) \left (\left (i \sqrt {c}\, \sqrt {a}\, \ln \left (x \right )^{2+k}-1\right ) \operatorname {hypergeom}\left (\left [\frac {\left (k +3\right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (4+2 k \right )}\right ], \left [\frac {k +3}{2+k}\right ], \frac {2 i \sqrt {c}\, \sqrt {a}\, \ln \left (x \right )^{2+k}}{2+k}\right )+i c_{3} \ln \left (x \right )^{k +1} \sqrt {a}\, \sqrt {c}\, \operatorname {hypergeom}\left (\left [\frac {\left (k +1\right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (4+2 k \right )}\right ], \left [\frac {k +1}{2+k}\right ], \frac {2 i \sqrt {c}\, \sqrt {a}\, \ln \left (x \right )^{2+k}}{2+k}\right )\right )\right )}{\left (k +3\right ) \left (k +1\right ) a \left (\operatorname {hypergeom}\left (\left [\frac {\left (k +1\right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (4+2 k \right )}\right ], \left [\frac {k +1}{2+k}\right ], \frac {2 i \sqrt {c}\, \sqrt {a}\, \ln \left (x \right )^{2+k}}{2+k}\right ) c_{3} +\operatorname {hypergeom}\left (\left [\frac {\left (k +3\right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (4+2 k \right )}\right ], \left [\frac {k +3}{2+k}\right ], \frac {2 i \sqrt {c}\, \sqrt {a}\, \ln \left (x \right )^{2+k}}{2+k}\right ) \ln \left (x \right )\right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime } x -a y^{2}=b \ln \left (x \right )^{k}+c \ln \left (x \right )^{2 k +2} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {a y^{2}+b \ln \left (x \right )^{k}+c \ln \left (x \right )^{2 k +2}}{x} \end {array} \]
Maple trace
`Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear trying Bernoulli trying separable trying inverse linear trying homogeneous types: trying Chini differential order: 1; looking for linear symmetries trying exact Looking for potential symmetries trying Riccati trying Riccati sub-methods: trying Riccati_symmetries trying Riccati to 2nd Order -> Calling odsolve with the ODE`, diff(diff(y(x), x), x) = -(diff(y(x), x))/x-a*(b*ln(x)^k+c*ln(x)^(2+2*k))*y(x)/x^2, y(x)` Methods for second order ODEs: --- Trying classification methods --- trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) -> Trying changes of variables to rationalize or make the ODE simpler trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Trying an equivalence, under non-integer power transformations, to LODEs admitting Liouvillian solutions. -> Trying a Liouvillian solution using Kovacics algorithm <- No Liouvillian solutions exists -> Trying a solution in terms of special functions: -> Bessel -> elliptic -> Legendre -> Whittaker -> hyper3: Equivalence to 1F1 under a power @ Moebius <- hyper3 successful: received ODE is equivalent to the 1F1 ODE <- Whittaker successful <- special function solution successful Change of variables used: [x = exp(t)] Linear ODE actually solved: a*(b*t^k+c*t^(2+2*k))*u(t)+diff(diff(u(t),t),t) = 0 <- change of variables successful <- Riccati to 2nd Order successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 480
dsolve(x*diff(y(x),x)=a*y(x)^2+b*(ln(x))^k+c*(ln(x))^(2*k+2),y(x), singsol=all)
\[ y \left (x \right ) = \frac {-\ln \left (x \right )^{1+k} \left (k +3\right ) \left (i \sqrt {c}\, \left (1+k \right ) \sqrt {a}-a b \right ) \operatorname {hypergeom}\left (\left [\frac {\left (3 k +5\right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (2 k +4\right )}\right ], \left [\frac {2 k +3}{k +2}\right ], \frac {2 i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}\right )+\left (-\left (i \left (k +3\right ) \sqrt {c}\, \sqrt {a}-a b \right ) \ln \left (x \right )^{k +2} c_{1} \operatorname {hypergeom}\left (\left [\frac {\left (3 k +7\right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (2 k +4\right )}\right ], \left [\frac {2 k +5}{k +2}\right ], \frac {2 i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}\right )+\left (k +3\right ) \left (\left (i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}-1\right ) c_{1} \operatorname {hypergeom}\left (\left [\frac {\left (k +3\right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (2 k +4\right )}\right ], \left [\frac {k +3}{k +2}\right ], \frac {2 i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}\right )+i \ln \left (x \right )^{1+k} \sqrt {a}\, \sqrt {c}\, \operatorname {hypergeom}\left (\left [\frac {\left (1+k \right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (2 k +4\right )}\right ], \left [\frac {1+k}{k +2}\right ], \frac {2 i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}\right )\right )\right ) \left (1+k \right )}{\left (\operatorname {hypergeom}\left (\left [\frac {\left (k +3\right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (2 k +4\right )}\right ], \left [\frac {k +3}{k +2}\right ], \frac {2 i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}\right ) \ln \left (x \right ) c_{1} +\operatorname {hypergeom}\left (\left [\frac {\left (1+k \right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (2 k +4\right )}\right ], \left [\frac {1+k}{k +2}\right ], \frac {2 i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}\right )\right ) a \left (k +3\right ) \left (1+k \right )} \]
✓ Solution by Mathematica
Time used: 3.775 (sec). Leaf size: 807
DSolve[x*y'[x]==a*y[x]^2+b*(Log[x])^k+c*(Log[x])^(2*k+2),y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -\frac {\log ^{k+1}(x) \left (\sqrt {c} c_1 (k+2) \sqrt {-(k+2)^2} \operatorname {HypergeometricU}\left (\frac {1}{2} \left (\frac {\sqrt {a} b}{\sqrt {c} \sqrt {-(k+2)^2}}+\frac {k+1}{k+2}\right ),\frac {k+1}{k+2},\frac {2 \sqrt {a} \sqrt {c} \log ^{k+2}(x)}{\sqrt {-(k+2)^2}}\right )+c_1 \left (\sqrt {a} b (k+2)+\sqrt {c} \sqrt {-(k+2)^2} (k+1)\right ) \operatorname {HypergeometricU}\left (\frac {1}{2} \left (\frac {\sqrt {a} b}{\sqrt {c} \sqrt {-(k+2)^2}}+\frac {3 k+5}{k+2}\right ),\frac {2 k+3}{k+2},\frac {2 \sqrt {a} \sqrt {c} \log ^{k+2}(x)}{\sqrt {-(k+2)^2}}\right )+\sqrt {c} (k+2) \sqrt {-(k+2)^2} \left (L_{-\frac {\sqrt {a} b}{2 \sqrt {c} \sqrt {-(k+2)^2}}-\frac {k+1}{2 k+4}}^{-\frac {1}{k+2}}\left (\frac {2 \sqrt {a} \sqrt {c} \log ^{k+2}(x)}{\sqrt {-(k+2)^2}}\right )+2 L_{-\frac {\sqrt {a} b}{2 \sqrt {c} \sqrt {-(k+2)^2}}-\frac {3 k+5}{2 k+4}}^{\frac {k+1}{k+2}}\left (\frac {2 \sqrt {a} \sqrt {c} \log ^{k+2}(x)}{\sqrt {-(k+2)^2}}\right )\right )\right )}{\sqrt {a} (k+2)^2 \left (L_{-\frac {\sqrt {a} b}{2 \sqrt {c} \sqrt {-(k+2)^2}}-\frac {k+1}{2 k+4}}^{-\frac {1}{k+2}}\left (\frac {2 \sqrt {a} \sqrt {c} \log ^{k+2}(x)}{\sqrt {-(k+2)^2}}\right )+c_1 \operatorname {HypergeometricU}\left (\frac {1}{2} \left (\frac {\sqrt {a} b}{\sqrt {c} \sqrt {-(k+2)^2}}+\frac {k+1}{k+2}\right ),\frac {k+1}{k+2},\frac {2 \sqrt {a} \sqrt {c} \log ^{k+2}(x)}{\sqrt {-(k+2)^2}}\right )\right )} \\ y(x)\to \frac {\log ^{k+1}(x) \left (-\frac {\left (\sqrt {a} b (k+2)+\sqrt {c} \sqrt {-(k+2)^2} (k+1)\right ) \operatorname {HypergeometricU}\left (\frac {1}{2} \left (\frac {\sqrt {a} b}{\sqrt {c} \sqrt {-(k+2)^2}}+\frac {3 k+5}{k+2}\right ),\frac {2 k+3}{k+2},\frac {2 \sqrt {a} \sqrt {c} \log ^{k+2}(x)}{\sqrt {-(k+2)^2}}\right )}{\operatorname {HypergeometricU}\left (\frac {1}{2} \left (\frac {\sqrt {a} b}{\sqrt {c} \sqrt {-(k+2)^2}}+\frac {k+1}{k+2}\right ),\frac {k+1}{k+2},\frac {2 \sqrt {a} \sqrt {c} \log ^{k+2}(x)}{\sqrt {-(k+2)^2}}\right )}-\sqrt {c} \sqrt {-(k+2)^2} (k+2)\right )}{\sqrt {a} (k+2)^2} \\ \end{align*}